The probability of happening the event A is 0.7 and that of event B is 0.5. If the events are independent, what is the probability of of happening
a) either of them
b) none of them
I know its simple..but I am stuck
Thanks
a) Since the events are not mutually exclusive, P (A or B)= P(A) + P (B) - P (AUB) = 0.7+ 0.5 - 0.35=0.85
b) 1-0.85 = 0.15
Please discard the previous ans. instead of 0.7 i used 0.8 for P(A). Now I hope I am correct. am I?
P(A union B)=P(A)+P(B)-P(A intersection B) (see Probability of a union here)
but A and B are independent implies P(A intersection B)=P(A)*P(B)
As a consequence:
P(either happens)=P(A union B)=P(A)+P(B)-P(A intersection B)=P(A)+P(B)-P(A)*P(B)=0.7+0.5-0.7*0.5=0.85
and
P(none happens)=1-P(either happens)=1-0.85=0.15